Answer to Question #30682 in Other Chemistry for Kel

If you have 84.0 grams of N2 and 12.0 grams of H2 for the reaction shown below, what is the liminting reactant?N2 (g) + 3H2 (g)------&gt; 2NH3 (g)

Expert's answer

2013-05-21T09:58:04-0400

According to this reaction we will calculate what amount of H2 is being needed for 84.0 grams of N2.n = m/Mwhere n - amount of molesm - massM - molar massn(N2) = 84/24 = 3.5 molesn(H2) = 3*n(N2) = 3*3.5 = 10.5 molesSo the needed amount of H2 is 10.5 moles. The real amount of H2 in 12 grams is:n(H2) = 12/1 = 12 molesThe needed amount of H2 is greater than the real one. So the amount of this element is limiting.

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